JEE Main 2018 (16 April Shift 1)

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Animated Solution for Mathematics - Statistics: The mean and the standard deviation(s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

Q: Can I calculate the new standard deviation using $\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}$ instead of the shortcut formula?

Absolutely! The shortcut formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is just a mathematical convenience. If you prefer the definition, calculate the deviations from the new mean (10): $(9-10)^2, (9-10)^2, (9-10)^2, (9-10)^2, (14-10)^2$. Sum these, divide by 5, and take the square root. You will arrive at the exact same answer. Mathematics is consistent!

Select Answer:

Visualized Solution

Initial State: $σ = 0$

Given: $n = 5$ , Mean $\overset{x}{ˉ} = 9$ , S.D. $σ = 0$ .

Implication of $σ = 0$

Property: If $σ = 0$ , then all observations are equal to the mean.
Initial observations: $x_{1} = x_{2} = x_{3} = x_{4} = x_{5} = 9$ .

Changing One Observation

New Mean $\overset{x}{ˉ}^{'} = 10$ .
Let the changed observation be $x_{5}^{'}$ .

Sum of New Observations

Sum of new observations: $\sum x_{i}^{'} = n \times \overset{x}{ˉ}^{'}$ .
$\sum x_{i}^{'} = 5 \times 10 = 50$ .

Finding the New Value $x_{5}^{'}$

Equation: $9 + 9 + 9 + 9 + x_{5}^{'} = 50$ .
$36 + x_{5}^{'} = 50 ⟹ x_{5}^{'} = 14$ .

Variance Formula Setup

Formula for Variance: $σ^{2} = \frac{\sum ( x _{i}^{'} ) ^{2}}{n} - (\overset{x}{ˉ}^{'})^{2}$ .

Substituting Values

Substitute values into the variance formula.
$σ^{2} = \frac{9 ^{2} + 9 ^{2} + 9 ^{2} + 9 ^{2} + 1 4 ^{2}}{5} - 1 0^{2}$ .

Computing Squares

Calculate the squares of the observations.
$9^{2} = 81$ and $1 4^{2} = 196$ .
Numerator part: $4 \times 81 + 196$ .

Computing Sum of Squares

Compute the total sum of squares.
$4 \times 81 = 324$ .
$\sum (x_{i}^{'})^{2} = 324 + 196 = 520$ .

Computing Variance $σ^{2}$

$σ^{2} = \frac{520}{5} - 100$ .
$σ^{2} = 104 - 100 = 4$ .

Final Standard Deviation $σ$

Standard Deviation $σ = Variance$ .
$σ = 4 = 2$ .
Final Answer: 2

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The Sigma Insight: Variance and Standard Deviation

Solution Diagram

Analyzing the Stillness of Zero

In statistics, a standard deviation of zero represents a state of absolute uniformity. When the standard deviation of five observations is

σ = 0

, it implies there is no spread in the data.

Every single data point must be identical to the mean. Given the mean is

9

, every observation

x_{1}, x_{2}, x_{3}, x_{4}, x_{5}

must be equal to

9

The Perturbation

A Shift in Balance

We introduce a change by replacing one of the nines with a new value,

x_{5}^{'}

. This shift changes the mean of the system to

\overset{x}{ˉ}^{'} = 10

The mean is defined by the relationship:

\overset{x}{ˉ} = \frac{\sum x _{i}}{n}

Since the number of observations

n

remains

5

and the new mean is

10

, the new sum of observations must be:

\sum x_{i} = 5 \times 10 = 50

We set up the equation for the new sum using the four original values and the unknown

x_{5}^{'}

9 + 9 + 9 + 9 + x_{5}^{'} = 50

36 + x_{5}^{'} = 50

x_{5}^{'} = 14

Rebuilding the Variance

With our new set of observations—four nines and one fourteen—we calculate the new variance using the computational formula:

σ^{2} = \frac{\sum ( x _{i}^{'} ) ^{2}}{n} - (\overset{x}{ˉ}^{'})^{2}

First, we calculate the sum of the squares of the new observations:

\sum (x_{i}^{'})^{2} = 9^{2} + 9^{2} + 9^{2} + 9^{2} + 1 4^{2}

\sum (x_{i}^{'})^{2} = 4 \times 81 + 196 = 324 + 196 = 520

Now, we substitute these values into the variance formula:

σ^{2} = \frac{520}{5} - (10)^{2}

σ^{2} = 104 - 100

σ^{2} = 4

The Final Reveal

The standard deviation is the square root of the variance. Therefore, we calculate:

σ = 4 = 2

The final standard deviation of the new set of observations is $2$ . This result demonstrates how a single change in a data set propagates through the mean to fundamentally alter the spread of the distribution.

Common Traps & FAQs

Wait, why does

σ = 0

mean all observations are equal? Doesn't it just mean they are close?

Standard deviation is defined as the square root of the variance, where variance is the average of the squared differences from the mean:

σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2}

. If

σ = 0

, then

\sum (x_{i} - \overset{x}{ˉ})^{2} = 0

. Since squares of real numbers are always non-negative, the only way their sum can be zero is if each individual term

(x_{i} - \overset{x}{ˉ})^{2}

is zero. This forces

x_{i} = \overset{x}{ˉ}

for every

i

Can I calculate the new standard deviation using

σ = \frac{\sum ( x _{i} - x ˉ ) ^{2}}{n}

instead of the shortcut formula?

Absolutely! The shortcut formula

σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2}

is just a mathematical convenience. If you prefer the definition, calculate the deviations from the new mean (10):

(9 - 10)^{2}, (9 - 10)^{2}, (9 - 10)^{2}, (9 - 10)^{2}, (14 - 10)^{2}

. Sum these, divide by 5, and take the square root. You will arrive at the exact same answer. Mathematics is consistent!

Is it possible to have a negative standard deviation?

Never! By definition, standard deviation is the square root of variance, and variance is a sum of squared deviations. Since squares are always positive or zero, the standard deviation must always be

\geq 0

. If you ever get a negative value, it's a sign to recheck your arithmetic.

Animated Solution for Mathematics - Statistics: The mean and the standard deviation(s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

Select Answer:

Visualized Solution

Initial State: $σ = 0$

Implication of $σ = 0$

Changing One Observation

Sum of New Observations

Finding the New Value $x_{5}^{'}$

Variance Formula Setup

Substituting Values

Computing Squares

Computing Sum of Squares

Computing Variance $σ^{2}$

Final Standard Deviation $σ$

The Sigma Insight: Variance and Standard Deviation

Analyzing the Stillness of Zero

The Perturbation

Rebuilding the Variance

The Final Reveal

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Animated Solution for Mathematics - Statistics: The mean and the standard deviation(s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

Select Answer:

Visualized Solution

Initial State: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ=0

Implication of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ=0

Changing One Observation

Sum of New Observations

Finding the New Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x5′​

Variance Formula Setup

Substituting Values

Computing Squares

Computing Sum of Squares

Computing Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ2

Final Standard Deviation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ

The Sigma Insight: Variance and Standard Deviation

Analyzing the Stillness of Zero

The Perturbation

Rebuilding the Variance

The Final Reveal

Similar Questions

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑i=19​(xi​−5)=9 and ∑i=19​(xi​−5)2=45, then the standard deviation of the 9 items x1​,x2​,…,x9​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A data consists of n observations: x1​,x2​,.....,xn​. If ∑i=1n​(xi​+1)2=9n and ∑i=1n​(xi​−1)2=5n, then the standard deviation of this data is :

Initial State: $σ = 0$

Implication of $σ = 0$

Finding the New Value $x_{5}^{'}$

Computing Variance $σ^{2}$

Final Standard Deviation $σ$

If $\sum_{i = 1}^{9} (x_{i} - 5) = 9$ and $\sum_{i = 1}^{9} (x_{i} - 5)^{2} = 45$ , then the standard deviation of the 9 items $x_{1}, x_{2}, \dots, x_{9}$ is :

A data consists of $n$ observations: $x_{1}, x_{2}, ....., x_{n}$ . If $\sum_{i = 1}^{n} (x_{i} + 1)^{2} = 9 n$ and $\sum_{i = 1}^{n} (x_{i} - 1)^{2} = 5 n$ , then the standard deviation of this data is :